Diskreetne matemaatika kodutöö
3.
f ( x1 , x 2 , x3 , x 4 ) = ( x 2 x3 x 4 ) & ( x1 x3 x4 ) & ( x1 x3 ) & ( x1 x 2 ) =
= ( x1 x 2 x 2 x3 x 2 x 4 x1 x3 x3 x 4 x1 x 4 x 4 x3 ) &
& ( x1 x3 x 2 ) = ( x1 x1 x2 x1 x2 x3 x1 x2 x 4 x1 x1 x3 x1 x3 x 4 x1 x1 x 4 x1 x4 x3
x1 x 2 x3 x 2 x 2 x3 x3 x 2 x 2 x4 x3 x2 x1 x3 x3 x 2 x3 x 4 x3 x 2 x1 x4 x3 x2 x4 x3 x3 x2 ) =
= ( x1 x 2 x3 x1 x 2 x 4 x1 x3 x 4 x1 x4 x3 x1 x3 x2 x4 x3 x2 )
Ei lange kokku punktis 2 leitud MDNKga. Koostan tõeväärtutabelid, et kontrollida
loogilist võrdsust.
MDNK
f ( x1 x2 x3 x4 ) = x1 x2 x1 x3 x4 x1 x2 x3 x3 x4
x1 , x2 , x3 , x4 x1 x2 x1 x3 x4 x1 x2 x3 x3 x4 f ( x1 x2 x3 x4 )
0000 0 0 0 0 0
0001 0 1 0 0 1
0010 0 0 0 1 1
0011 0 0 0 0 0